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Description: A different way of using double 2 angles : If we get the angle around the 3rd axis giving: x^2-y^2 2*x*y ?? And double the angle around the x axis using: ?? y^2-z^2 2*y*z So if we take 2*x*y as the given "correct" new value for y then we can calculate the new z value such that both angles are maintained : x^2 -y^2 2*x*y 4*x*y^2*z/(y^2-z^2) Of course this gives a slight issue when the new angle around the x axis is 90 degrees i.e. when (y^2-z^2) is zero, but in such a case I've simply used 2*y*z as the new z value because in order to maintain the correct magnitude we have to fudge that anyway. Stats: Total Favorities: 0 View Who Favorited Filesize: 944.01kB Height: 960 Width: 1280 Discussion Topic: View Topic Keywords: two angles Posted by: David Makin July 09, 2014, 03:38:52 AM Rating: by 1 members. Image Linking Codes
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David Makin | July 09, 2014, 03:41:18 AM Oops - the above is correct for doubling the angles, but I should have pointed out this is z^8+c using the above triplex method so of course both angles are multiplied by 8 in this case. |
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